Question

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times\) \(10^{-3} M ?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is \(1.62 \mathrm{~V}\). Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume \(\mathrm{Al}\) is oxidized.)

Short answer

The potential of the cell at \(25^{\circ} C\) with \(\left[\mathrm{Al}^{3+}\right]=7.2 \times 10^{-3} M\) is approximately 1.28V, and the concentration of \(\mathrm{Al^{3+}}\) in the unknown solution is approximately \(1.6 \times 10^{-4}M\).

Step-by-step solution

Write the half-cell reactions

Write the half-cell reactions for both \(\mathrm{Ni^{2+}}\) and \(\mathrm{Al^{3+}}\): Ni\(^{2+}\) + 2e\(^-\) → Ni, with E\(^o\) = -0.25V (reduction) Al\(^{3+}\) + 3e\(^-\) → Al, with E\(^o\) = -1.66V (reduction) Since aluminum is being oxidized, we need to reverse the aluminum half-cell reaction, which gives: Al → Al\(^{3+}\) + 3e\(^-\), with E\(^o\) = 1.66V (oxidation)

Determine the overall cell reaction

Combine the two half-cell reactions and obtain the overall cell reaction as follows: Ni\(^{2+}\) + 2Al → Ni + 2Al\(^{3+}\)

Calculate the standard cell potential

We can now calculate the standard cell potential, E\(^o_{cell}\) as the difference between standard reduction potentials: E\(^o_{cell}\) = E\(^o_{oxidation}\) - E\(^o_{reduction}\) = 1.66V - (-0.25V) = 1.91V

Calculate the cell potential using the Nernst equation

Given T = \(25^{\circ} C = 298K\) and the \(\mathrm{Al^{3+}}\) concentration in the problem, calculate the cell potential using the Nernst equation: \(E_{cell} = E_{cell}^o - \frac{RT}{nF} \ln Q\) \(E_{cell} = 1.91V - \frac{8.314 J mol^{-1} K^{-1} \times 298K}{(2 \times 3)e C mol^{-1}} \ln \frac{[\mathrm{Al^{3+}}]^2}{[\mathrm{Ni^{2+}}]}\) \(E_{cell} = 1.91V - \frac{8.314 \times 298}{(2 \times 3) \times 96,485} \ln \frac{(7.2 \times 10^{-3})^2}{1}\) Solving this equation yields: \(E_{cell} \approx 1.28V\) So, the potential of the cell at \(25^{\circ} C\) with \(\left[\mathrm{Al}^{3+}\right]=7.2 \times 10^{-3} M\) is approximately 1.28V. #b. Calculating the Concentration of Al\(^{3+}\) Based on the Measured Cell Potential#

Solve the Nernst equation for concentration

Since we are given the cell potential, E\(_{cell}\) = 1.62V, and we have already calculated the standard cell potential, E\(^o_{cell}\) = 1.91V, we can solve the Nernst equation for the unknown concentration, \(\left[\mathrm{Al}^{3+}\right]\): From the Nernst equation: \(1.62V = 1.91V - \frac{8.314 \times 298}{(2 \times 3) \times 96,485} \ln \frac{[\mathrm{Al^{3+}}]^2}{[\mathrm{Ni^{2+}}]}\) Moving the constant part of the equation over to the other side and then solving for the logarithm yields: \(\ln\frac{[\mathrm{Al^{3+}}]^2}{[\mathrm{Ni^{2+}}]} = \frac{(1.91V - 1.62V) \times (2 \times 3) \times 96,485}{8.314 \times 298}\)

Calculate the concentration of Al\(^{3+}\) from the logarithmic part

Now that we have isolated the logarithmic part, we can take the exponential of both sides and then solve for \(\left[\mathrm{Al}^{3+}\right]\): \(\frac{[\mathrm{Al^{3+}}]^2}{[\mathrm{Ni^{2+}}]} = e^{\frac{(1.91V - 1.62V) \times (2 \times 3) \times 96,485}{8.314 \times 298}}\) Multiplying both sides of the equation by \(\mathrm{Ni^{2+}}\), and then taking the square root: \([\mathrm{Al^{3+}}] = \sqrt{\mathrm{Ni^{2+}} \times e^{\frac{(1.91V - 1.62V) \times (2 \times 3) \times 96,485}{8.314 \times 298}}}\) Substituting \(\mathrm{Ni^{2+}}=1M\) and calculating the value, we get: \([\mathrm{Al^{3+}}] \approx 1.6 \times 10^{-4}\mathrm{M}\) Hence, the concentration of \(\mathrm{Al^{3+}}\) in the unknown solution is approximately \(1.6 \times 10^{-4}M\).

Key concepts

Understanding the Nernst Equation

The Nernst equation is a fundamental principle in electrochemistry that describes how the voltage of an electrochemical cell is affected by the concentration of the ions involved in the chemical reaction. It allows us to predict the cell potential under non-standard conditions, which is key to understanding batteries, corrosion, and various industrial processes.

At its core, the equation accounts for the thermal energy available to the system (through the temperature, T) and the concentrations of reactants and products (through the reaction quotient, Q). The equation is mathematically expressed as:
\[\begin{equation}E = E^0 - \frac{RT}{nF} \ln Q\text{}\end{equation}\]where:

• \(E\) is the cell potential under non-standard conditions,
• \(E^0\) is the standard cell potential,
• \(R\) is the ideal gas constant (8.314 J mol⁻¹ K⁻¹),
• \(T\) is the temperature in Kelvin,
• \(n\) is the number of moles of electrons transferred in the reaction,
• \(F\) is Faraday's constant (96,485 C mol⁻¹),
• \(Q\) is the reaction quotient, a ratio of product activities to reactant activities for the reaction as it proceeds to equilibrium.

The Nernst equation is routinely used to calculate how the cell potential changes as the concentration of the chemical species vary, which was demonstrated in the exercise solution when adjusting the standard cell potential with changing aluminum ion concentration.

Deciphering Half-Cell Reactions

Half-cell reactions are at the heart of electrochemistry. They are the individual reactions that occur at each electrode in an electrochemical cell. In essence, every electrochemical reaction can be broken down into two half-reactions: one for the oxidation process and one for the reduction process. These reactions are crucial for understanding how electrons are transferred during a chemical reaction.

In the textbook solution, the half-cell reactions for nickel and aluminum were written as:

• Ni²⁺ + 2e⁻ → Ni (reduction),
• Al ³⁺ + 3e⁻ → Al (oxidation - since Al is being oxidized, this reaction is reversed).

These reactions show that nickel ions gain electrons (reduce), and aluminum loses electrons (oxidize). In an electrochemical cell, these reactions occur simultaneously, one at the anode and the other at the cathode. The flow of electrons from one reaction to the other through an external circuit is what generates electric current, a principle that's fundamental to the workings of batteries and other electrochemical cells.

Explaining Standard Reduction Potential

The standard reduction potential (\(E^0\)) is a measure of the tendency of a chemical species to gain electrons and thereby be reduced. It is determined under standard conditions: a 1 Molar concentration of ions, a pressure of 1 atmosphere for gases, and a temperature of 25°C or 298 K. Standard reduction potentials are used as references to predict the direction of an electrochemical reaction and calculate the standard cell potential.

When combining two half-cell reactions to form the overall cell reaction, as seen in the exercise, you compare the standard reduction potentials of each half-cell. The substance with the higher reduction potential will undergo reduction, serving as the cathode, and the one with the lower will be oxidized, becoming the anode. The difference in their standard reduction potentials gives the standard cell potential (\(E^0_{cell}\)), representing the maximum voltage the cell can produce under standard conditions. If either half-cell is reversed (as was the case with the aluminum half-cell in the exercise), its sign is changed, turning a reduction potential into an oxidation potential. Calculating the standard cell potential is crucial; once you have this value, the Nernst equation can adjust it to reflect real-world conditions, including non-standard concentrations.